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Prove that:$ \frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{1}{\sec \theta-\tan \theta} $
To do:
We have to prove that \( \frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{1}{\sec \theta-\tan \theta} \).
Solution:
We know that,
$\sin^2 A+\cos^2 A=1$
$\operatorname{cosec}^2 A-\cot^2 A=1$
$\sec^2 A-\tan^2 A=1$
$\cot A=\frac{\cos A}{\sin A}$
$\tan A=\frac{\sin A}{\cos A}$
$\operatorname{cosec} A=\frac{1}{\sin A}$
$\sec A=\frac{1}{\cos A}$
Therefore,
$\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{\tan \theta-1+\sec \theta}{\tan \theta+1-\sec \theta}$ (Dividing numerator and denominator by $\cos \theta$ )
$=\frac{(\tan \theta+\sec \theta)-1}{(\tan \theta-\sec \theta)+1}$
Multiplying and dividing by $(\tan \theta-\sec \theta)$, we get,
$=\frac{(\tan \theta+\sec \theta-1)(\tan \theta-\sec \theta)}{(\tan \theta-\sec \theta+1)(\tan \theta-\sec \theta)}$
$=\frac{\left(\tan ^{2} \theta-\sec ^{2} \theta\right)-(\tan \theta-\sec \theta)}{(\tan \theta-\sec \theta+1)(\tan \theta-\sec \theta)}$
$=\frac{-1-\tan \theta+\sec \theta}{(\tan \theta-\sec \theta+1)(\tan \theta-\sec \theta)}$
$=\frac{-(\tan \theta-\sec \theta+1)}{(\tan \theta-\sec \theta+1)(\tan \theta-\sec \theta)}$
$=\frac{-1}{\tan \theta-\sec \theta}$
$=\frac{1}{\sec \theta-\tan \theta}$
Hence proved.
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